Given two arrays count all pairs whose sum is an odd number

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Given two arrays of N and M integers. The task is to find the number of unordered pairs formed of elements from both arrays in such a way that their sum is an odd number.
Note: An element can only be one pair.
Examples:

Input: a[] = {9, 14, 6, 2, 11}, b[] = {8, 4, 7, 20}
Output: 3
{9, 20}, {14, 7} and {11, 8}
Input: a[] = {2, 4, 6}, b[] = {8, 10, 12}
Output: 0

Approach: Count the number of odd and even numbers in both the arrays and the answer to the number of pairs will be min(odd1, even2) + min(odd2, even1), because odd + even is only odd.
Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns the number of pairs
int count_pairs(int a[], int b[], int n, int m)
{
 
    // Count of odd and even numbers
    int odd1 = 0, even1 = 0;
    int odd2 = 0, even2 = 0;
 
    // Traverse in the first array
    // and count the number of odd
    // and even numbers in them
    for (int i = 0; i < n; i++) {
        if (a[i] % 2)
            odd1++;
        else
            even1++;
    }
 
    // Traverse in the second array
    // and count the number of odd
    // and even numbers in them
    for (int i = 0; i < m; i++) {
        if (b[i] % 2)
            odd2++;
        else
            even2++;
    }
 
    // Count the number of pairs
    int pairs = min(odd1, even2) + min(odd2, even1);
 
    // Return the number of pairs
    return pairs;
}
 
// Driver code
int main()
{
    int a[] = { 9, 14, 6, 2, 11 };
    int b[] = { 8, 4, 7, 20 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    cout << count_pairs(a, b, n, m);
 
    return 0;
}

Java

// Java program to implement
// the above approach
 
class GFG {
 
    // Function that returns the number of pairs
    static int count_pairs(int a[], int b[], int n, int m)
    {
 
        // Count of odd and even numbers
        int odd1 = 0, even1 = 0;
        int odd2 = 0, even2 = 0;
 
        // Traverse in the first array
        // and count the number of odd
        // and even numbers in them
        for (int i = 0; i < n; i++) {
            if (a[i] % 2 == 1) {
                odd1++;
            }
            else {
                even1++;
            }
        }
 
        // Traverse in the second array
        // and count the number of odd
        // and even numbers in them
        for (int i = 0; i < m; i++) {
            if (b[i] % 2 == 1) {
                odd2++;
            }
            else {
                even2++;
            }
        }
 
        // Count the number of pairs
        int pairs = Math.min(odd1, even2) + Math.min(odd2, even1);
 
        // Return the number of pairs
        return pairs;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 9, 14, 6, 2, 11 };
        int b[] = { 8, 4, 7, 20 };
        int n = a.length;
        int m = b.length;
        System.out.println(count_pairs(a, b, n, m));
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python 3 program to implement
# the above approach
 
# Function that returns 
# the number of pairs
def count_pairs(a, b, n, m):
     
    # Count of odd and even numbers
    odd1 = 0
    even1 = 0
    odd2 = 0
    even2 = 0
 
    # Traverse in the first array
    # and count the number of odd
    # and even numbers in them
    for i in range(n):
        if (a[i] % 2):
            odd1 += 1
        else:
            even1 += 1
 
    # Traverse in the second array
    # and count the number of odd
    # and even numbers in them
    for i in range(m):
        if (b[i] % 2):
            odd2 += 1
        else:
            even2 += 1
 
    # Count the number of pairs
    pairs = (min(odd1, even2) +
             min(odd2, even1))
 
    # Return the number of pairs
    return pairs
 
# Driver code
if __name__ == '__main__':
    a = [9, 14, 6, 2, 11]
    b = [8, 4, 7, 20]
    n = len(a)
    m = len(b)
    print(count_pairs(a, b, n, m))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement
// the above approach
using System;
 
class GFG {
 
    // Function that returns the number of pairs
    static int count_pairs(int[] a, int[] b, int n, int m)
    {
 
        // Count of odd and even numbers
        int odd1 = 0, even1 = 0;
        int odd2 = 0, even2 = 0;
 
        // Traverse in the first array
        // and count the number of odd
        // and even numbers in them
        for (int i = 0; i < n; i++) {
            if (a[i] % 2 == 1) {
                odd1++;
            }
            else {
                even1++;
            }
        }
 
        // Traverse in the second array
        // and count the number of odd
        // and even numbers in them
        for (int i = 0; i < m; i++) {
            if (b[i] % 2 == 1) {
                odd2++;
            }
            else {
                even2++;
            }
        }
 
        // Count the number of pairs
        int pairs = Math.Min(odd1, even2) + Math.Min(odd2, even1);
 
        // Return the number of pairs
        return pairs;
    }
 
    // Driver code
    static public void Main()
    {
        int[] a = { 9, 14, 6, 2, 11 };
        int[] b = { 8, 4, 7, 20 };
        int n = a.Length;
        int m = b.Length;
        Console.WriteLine(count_pairs(a, b, n, m));
    }
}
 
// This code contributed by ajit.

PHP

<?php
// PHP program to implement 
// the above approach 
 
// Function that returns the number of pairs 
function count_pairs($a, $b, $n, $m) 
{ 
    // Count of odd and even numbers 
    $odd1 = 0; $even1 = 0; 
    $odd2 = 0; $even2 = 0; 
 
    // Traverse in the first array 
    // and count the number of odd 
    // and even numbers in them 
    for ($i = 0; $i < $n; $i++)
    { 
        if ($a[$i] % 2) 
            $odd1++; 
        else
            $even1++; 
    } 
 
    // Traverse in the second array 
    // and count the number of odd 
    // and even numbers in them 
    for ($i = 0; $i < $m; $i++)
    { 
        if ($b[$i] % 2) 
            $odd2++; 
        else
            $even2++; 
    } 
 
    // Count the number of pairs 
    $pairs = min($odd1, $even2) + min($odd2, $even1); 
 
    // Return the number of pairs 
    return $pairs; 
} 
 
    // Driver code 
    $a = array( 9, 14, 6, 2, 11 ); 
    $b = array( 8, 4, 7, 20 ); 
    $n = count($a) ; 
    $m = count($b) ; 
     
    echo count_pairs($a, $b, $n, $m); 
     
    // This code is contributed by Ryuga
?>

Javascript

<script>
 
// JavaScript program to implement 
// the above approach 
 
// Function that returns the number of pairs 
function count_pairs(a, b, n, m) 
{ 
 
    // Count of odd and even numbers 
    let odd1 = 0, even1 = 0; 
    let odd2 = 0, even2 = 0; 
 
    // Traverse in the first array 
    // and count the number of odd 
    // and even numbers in them 
    for (let i = 0; i < n; i++) { 
        if (a[i] % 2) 
            odd1++; 
        else
            even1++; 
    } 
 
    // Traverse in the second array 
    // and count the number of odd 
    // and even numbers in them 
    for (let i = 0; i < m; i++) { 
        if (b[i] % 2) 
            odd2++; 
        else
            even2++; 
    } 
 
    // Count the number of pairs 
    let pairs = Math.min(odd1, even2) + Math.min(odd2, even1); 
 
    // Return the number of pairs 
    return pairs; 
} 
 
// Driver code 
    let a = [ 9, 14, 6, 2, 11 ]; 
    let b = [ 8, 4, 7, 20 ]; 
    let n = a.length; 
    let m = b.length; 
    document.write(count_pairs(a, b, n, m)); 
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output:

Time Complexity: O(n + m)

Auxiliary Space: O(1)

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